\(\sin^2a+cos^2a=1\Rightarrow sin^2a=1-0,8^2=0,36\)độ 0<=sina<=1 nên ta có \(sina=0.6\)

lại có \(\frac{sina}{cosa}=tana\Rightarrow tana=\frac{0,6}{0,8}=0.75\)

\(\frac{cosa}{sina}=cotga\Rightarrow cotga=\frac{0.8}{0.6}=\frac{4}{3}\)

\(cos^2\alpha=1-sin^2\alpha=1-\left(0,8\right)^2=0,36\)

\(\Rightarrow cos\alpha=0,6\)

\(1+cot^2\alpha=\dfrac{1}{sin^2\alpha}\Rightarrow cot^2\alpha=\dfrac{1}{sin^2\alpha}-1=\dfrac{9}{16}\)

\(\Rightarrow cot\alpha=0,75\)

\(tan\alpha=\dfrac{1}{cot\alpha}=\dfrac{1}{0,75}=\dfrac{4}{3}\)

HD để bạn tự lm cho dễ hiểu nhâ

-Dựa vào công thức sin^2a+cos^2a=1

=>cosa=?

-tana=sina/cosa

-cota=cosa/sina

\(sin^{2}x+cos^{2}x=1\)

\(\Leftrightarrow\)\(cos^{2}x=1-sin^{2}x\)

\(\Leftrightarrow\)\(cos^{2}x=1-(0,8)^{2}\)

\(cos^{}x=0,6\)

\(tanx=\dfrac{sinx}{cosx}=\dfrac{0,8}{0,6}=\dfrac{4}{3} \)

\(cotx=\dfrac{cosx}{sinx}=\dfrac{0,6}{0,8}=\dfrac{3}{4}\)

1: 

a: sin a=căn 3/2

\(cosa=\sqrt{1-sin^2a}=\sqrt{1-\dfrac{3}{4}}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\)

\(tana=\dfrac{\sqrt{3}}{2}:\dfrac{1}{2}=\sqrt{3}\)

cot a=1/tan a=1/căn 3

b: \(tana=2\)

=>cot a=1/tan a=1/2

\(1+tan^2a=\dfrac{1}{cos^2a}\)

=>\(\dfrac{1}{cos^2a}=5\)

=>cos^2a=1/5

=>cosa=1/căn 5

\(sina=\sqrt{1-cos^2a}=\sqrt{\dfrac{4}{5}}=\dfrac{2}{\sqrt{5}}\)

c: \(cosa=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\)

tan a=5/13:12/13=5/12

cot a=1:5/12=12/5

Lớp 9 nên coi như các góc này đều nhọn

a.

\(cosa=\sqrt{1-sin^2a}=\dfrac{15}{17}\)

\(tana=\dfrac{sina}{cosa}=\dfrac{8}{15}\)

\(cota=\dfrac{1}{tana}=\dfrac{15}{8}\)

b.

\(1+cot^2a=\dfrac{1}{sin^2a}\Rightarrow sina=\dfrac{1}{\sqrt{1+cot^2a}}=\dfrac{4}{5}\)

\(cosa=\sqrt{1-sin^2a}=\dfrac{3}{5}\)

\(tana=\dfrac{1}{cota}=\dfrac{4}{3}\)

a) \(\cos=\sqrt{1-\sin^2}=\sqrt{1-\dfrac{64}{289}}=\dfrac{15}{17}\)

\(\tan=\dfrac{\sin}{\cos}=\dfrac{8}{17}:\dfrac{15}{17}=\dfrac{8}{15}\)

\(\cot=\dfrac{\cos}{\sin}=\dfrac{15}{17}:\dfrac{8}{17}=\dfrac{15}{8}\)

Ta có: \(\frac{cosa+sina}{cosa-sina}=\frac{\frac{cosa}{cosa}+\frac{sina}{cosa}}{\frac{cosa}{cosa}-\frac{sina}{cosa}}=\frac{1+tana}{1-tana}=\frac{1+\frac{1}{3}}{1-\frac{1}{3}}=2\)